Problem: The tangent line to the graph of function $h$ at the point $(-2,-4)$ passes through the point $(1,5)$. Find $h'(-2)$. $h'(-2)=$
Answer: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $h'(-2)$ gives the slope of the tangent line to the graph of $h$ where $x=-2$, which is the point $(-2,-4)$. We know this line passes through $(-2,-4)$, and we are also given that it passes through $(1,5)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{5-(-4)}{1-(-2)} \\\\ &=\dfrac{9}{3} \\\\ &=3 \end{aligned}$ In conclusion, $h'(-2)=3$.